Consider the integral $$ \int_{C}\frac{1}{z-z_0} dz$$ where $C$ is the unit circle. I am asked to evaluate this integral in the case where $|z_0| > 1, |z_0| <1,$ and $|z_0|=1$.
The case where $|z_0| > 1$ is simple: inside the unit circle the function is analytic and therefore $\oint_C 1/(z-z_0) = 0$.
In the case where $|z_0| < 1$ is not too bad either. By hypothesis our circle is strictly inside the unit circle and hence we can construct a circle of finite radius $R$ that is completely inside the unit circle - call it $C_2$. Then I substituted $z - z_0 = Re^{i \theta}$ and got (utilizing the principle of deformation of contours): $$ \oint_{C_1}\frac{1}{z-z_0} dz = \oint_{C_2}\frac{1}{z-z_0} dz = \int_{0}^{2\pi}\frac{i Re^{i \theta}}{Re^{i \theta}} d\theta = 2 i \pi. $$
I'm pretty confident with how I solved the first 2 cases. The last case is where I am not so sure I am correct, and wish to get some advice. Here is what I argued:
Consider the case where $|z_0|$ < 1. As $z_0$ approaches the boundary of our unit circle, our choices of radii for $C_2$ get smaller and smaller. In the limit where $|z_0| \rightarrow 1$, $R \rightarrow 0$. Thus we take our previous integral and consider the limit that $R \rightarrow 0$:
$$ \lim_{R\rightarrow 0} \int_{0}^{2\pi}\frac{i Re^{i \theta}}{Re^{i \theta}} d\theta = \lim_{R \rightarrow 0} \int_{0}^{2\pi} i \, d\theta = 2 \pi i $$
and so when $|z_0| = 1$ our integral is still $2πi $.
The problem here is I feel like I can make the same argument from the other side… Consider the case where $|z_0| > 1$ and so $|z_0| = 1 + \epsilon$. For all $\epsilon > 0$ the value of the integral is 0. Thus, in the limit where $\epsilon \rightarrow 0$ the integral is equal to 0.
Does the fact that the limits don't agree imply that the integral is not defined for $|z_0| = 1$?
